Problem: $ f(x)=\sum_{n=0}^{\infty }{{{\left( -1 \right)}^{n+1}}\dfrac{2{{x}^{2n+1}}}{\left( 2n+1 \right)!}}$ $f'''(0)=$
Explanation: Note that because we are interested in what happens at $x=0$, we only need to pay attention to the constant term in the series that represents $f$. With that said, we now calculate derivatives. For clarity, we will work out one more term than we actually need to do the problem. $\begin{aligned} f'\left( x \right)&=-2+\dfrac{6{{x}^{2}}}{3!}-\dfrac{10x^4}{5!}+... \\\\ f''\left( x \right)&=\dfrac{12x}{3!}-\dfrac{40x^3}{5!}+... \\\\ f'''\left( x \right)&=\dfrac{12{^}}{3!}-\dfrac{120x^2}{5!}+... \end{aligned}$ So $f'''\left( 0 \right)=\dfrac{12}{3!}=2$.